## Loser66 one year ago Let A, B $$\in M(n, \mathbb C)$$and assume B is invertible. Show that 1) $$(B^{-1}AB)^k= B^{-1}A^kB$$ 2) Use 1 to show $$e^{tB^{-1}AB}=B^{-1}e^{tA}B$$ Please help

1. Loser66

1) is easy. But I don't know how to do 2 @ganeshie8 @dan815

2. ganeshie8

is it given that A is diagonal matrix ?

3. Loser66

Nope, that is all.

4. UsukiDoll

invertible meaning non-singular matrix.. there's an inverse.

5. Loser66

nvm, I got it. :) Thanks for being here.

6. Loser66

for 1) $$(B^{-1}AB)^k = (B^{-1}AB)(B^{-1}AB)........(B^{-1}AB)$$ open parenthesis and $$BB^{-1}=I$$ gives us the answer

7. Loser66

for 2) by definition $$e^{tA}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k$$

8. Loser66

we replace A by $$B^{-1}AB$$ then just simplify it to get the answer.

9. ganeshie8

$$e^{tB^{-1}AB}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}(B^{-1}AB)^k =\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB$$ ?

10. UsukiDoll

yeah.... this proof looks familiar. Miracrown helped me with one of them.

11. ganeshie8

how do you conclude

12. Loser66

@ganeshie8 sum and t, k are constants, right? hence, we can distribute inside, between $$B^{-1}$$ and A

13. ganeshie8

Are you saying we can take $$B^{-1}$$ out ?

14. Loser66

Yes

15. ganeshie8

$$\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B$$ ? this looks ok intuitively but idk what you're gonna use as justification

16. Loser66

Look at the definition of $$e^{tA}$$ above

17. ganeshie8

Yes looked.

18. Loser66

And we done, right?

19. ganeshie8

I am asking the justification for that distribution

20. ganeshie8

I think its not so obvious, at least for me it is not.

21. Loser66
22. dan815

|dw:1437277499951:dw|

23. dan815

|dw:1437277542960:dw|

24. dan815

|dw:1437277640041:dw|

25. dan815

therefore true

26. dan815

if u expand that summation u can show these 2 summations are equation therefore

27. dan815

|dw:1437277900211:dw|

28. Loser66

hey, barbecue, @ganeshie8 asked whether we can put the sum between B^-1 and A or not. You proved $$A(B_1+B_2) = AB_1 +AB_2$$ which is irrelevant to the question. ha!!

29. dan815

this is what we need

30. dan815

$\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B$

31. dan815

|dw:1437278072521:dw|

32. dan815

you see?

33. Loser66

Oh, yeah, barbecue!! Thanks for making it clear.

34. dan815

sure thing

35. anonymous

consider $$C=B^{-1}AB$$. now suppose $$C^{n-1}=B^{-1}A^{n-1}B$$. it follows that $$C^n=C\cdot C^{n-1}=B^{-1}ABB^{-1}A^{n-1}B=B^{-1}AA^{n-1}B=B^{-1}A^nB$$. so by induction it follows $$C^n=B^{-1}A^n B$$ define $$e^X=\sum_{n=0}^\infty \frac1{n!}X^n$$ in analogy with the real exponential. \begin{align*}e^{B^{-1}AB}=\sum_{n=0}^\infty\frac1{n!}(B^{-1}AB)&=\sum_{n=0}^\infty\frac1{n!}B^{-1}A^nB\\&=B^{-1}\sum_{n=0}^\infty\frac1{n!}A^n B\\&=B^{-1}\left(\sum_{n=0}^\infty\frac1{n!}A^n\right)B\\&=B^{-1}e^AB\end{align*}

36. anonymous

the above follows because matrix multiplication distributes over addition: $$B^{-1}(A_1+A_2)B=B^{-1}(A_1B+A_2B)=B^{-1}A_1B+B^{-1}A_2B$$